(1) When the friction force is zero, the resultant force of the supporting force and gravity provides the centripetal force, which is:
mgtanθ=mRsinθω02, the solution is
ω0=2gR
(2) When ω=(1+k)ω0, the combined force of gravity and supporting force is not enough to provide centripetal force, and the direction of friction force is downward along the tangent line of the tank wall.
According to Newton The second law is: fcos60°+Ncos30°=mRsin60°ω2.
fsin60°+mg=Nsin30°
Solving the two simultaneous equations, we get f=3k(2+k)2mg
When ω=(1-k ) When ω0, the friction force direction is upward along the tangent to the tank wall.
According to Newton’s second law, Ncos30°-fcos60°=mRsin60°ω2.
mg=Nsin30°+fsin60°
Solving the two equations simultaneously, we get f=3k(2?k)2mg.
Answer: (1) When the friction force on the small object is exactly zero, ω0 = 2gR.
(2) When ω=(1+k)ω0, the friction force direction is downward along the tangent line of the tank wall, and the magnitude is f=3k(2+k)2mg.
When ω=(1-k)ω0, the friction force direction is upward along the tangent line of the tank wall, and the magnitude is f=3k(2?k)2mg.